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Editorial Team
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Asked: 2026-06-09T00:29:34+00:00

Given an enum declared like so: enum { A, B, C, D }; What

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Given an enum declared like so:

enum {
    A,
    B,
    C,
    D
};

What is the general compiler support with reference to § 7.2 of the C++11 standard? Specifically, this excerpt from § 7.2.2:

If the first enumerator has no initializer, the value of the
corresponding constant is zero. An enumerator-definition without an
initializer gives the enumerator the value obtained by increasing the
value of the previous enumerator by one.

Can I expect the common, modern compilers (GCC, Intel, Clang, recent versions of CL, others), to give the same results, that is, A = 0, B = 1, C = 2, and D = 3?

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1 Answer

  1. Editorial Team

    The rule you cite is not new in C++11. It’s part of C++03, C++98, C11, C99, and C89. This rule existed before these languages were ever standardized. Java and C# both inherited this behavior with their enums.

    Yes, compilers support this part of the langauge. Just like they support if, switch, #define, ints and other basic language constructs.

    We’re not talking about r-value references or lambdas or something. This is core stuff from before many programmers today were even born.

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